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Sagot :
The vapour pressure of the solution of glycerol is 106.4 torr.
What is vapour pressure?
The vapour pressure is given as the pressure extered by the individual sample in a solution. According to Roult’s law, the vapour pressure of mixture is given as:
[tex]\rm Vapor\;pressure\;of\;solution=moles\;of\;solute\;\times\;vapour\;pressure\;of\;pure\;solvent[/tex]
The density of glycerol is 1.25 g/mL. The mass of glycerol in 261 mL will be:
[tex]\rm Density=\dfrac{mass}{volume}\\\\1.25\;g/mL=\dfrac{mass}{261\;mL}\\\\Mass =326.25\;g[/tex]
The moles of glycerol are given as:
[tex]\rm Moles=\dfrac{mass}{molar\;mass}\\\\Moles=\dfrac{326.25}{92.09\;g/mol}\\\\Moles=3.54\;moles[/tex]
The density of water is 1, thus the mass of 379 mL water is 379 grams. The moles of solvent are:
[tex]\rm Moles=\dfrac{379}{18\;g/mol}\\\\Moles=21.05\;moles[/tex]
The mole fraction of solute is given as:
[tex]\rm Mole \;fraction=\dfrac{moles\;of\;solute}{moles\;of\;solute\;+\;moles\;of\;solvent}\\\\Mole\;fraction=\dfrac{3.54}{3.54+21.05}\\\\Mole\;fraction=0.14[/tex]
The mole fraction of solute is 0.14. The vapour pressure of pure water at boiling point is equivalent to the atmospheric pressure i.e. 760 torr.
Substituting the values in Roult’s law, the vapour pressure of solution is given as:
[tex]\rm Vapor\;pressure=0.14\;\times\;760\;torr\\Vapor\;pressure=106.4\;torr[/tex]
The vapour pressure of the solution of glycerol is 106.4 torr.
Learn more about vapor pressure, here:
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